Question

Work out the gradient of the straight line that passes through (2,6) and (6,12)

Answer 1

Answer:

The gradient of the straight line that passes through (2, 6) and (6, 12) is $m = \frac{3}{2}$.

Step-by-step explanation:

Mathematically speaking, lines are represented by following first-order polynomials of the form:

$y = b + m\cdot x$ (1)

Where:

$x$ - Independent variable.

$y$ - Dependent variable.

$m$ - Slope.

$b$ - Intercept.

The gradient of the function is represented by the first derivative of the function:

$y' = m$

Then, we conclude that the gradient of the staight line is the slope. According to Euclidean Geometry, a line can be form after knowing the locations of two distinct points on plane. By definition of secant line, we calculate the slope:

$m = \frac{y_{B}-y_{A}}{x_{B}-x_{A}}$ (2)

Where:

$x_{A}$, $y_{A}$ - Coordinates of point A.

$x_{B}$, $y_{B}$ - Coordinates of point B.

If we know that $A(x,y) = (2,6)$ and $B(x,y) = (6,12)$, then the gradient of the straight line is:

$m = \frac{12-6}{6-2}$

$m = \frac{6}{4}$

$m = \frac{3}{2}$

The gradient of the straight line that passes through (2, 6) and (6, 12) is $m = \frac{3}{2}$.